换元积分法这是我的做法∵d( sin(x/2) ) = cos(x/2)/2 dx∴cos(x/2)dx = 2d( sin(x/2) )∴原式=∫ 2cosx d( sin(x/2) ) =2∫ 1-2sin^2(x/2) d( (sin(x/2) ) =2sin(x/2) - 2/3sin^3(x/2) + C 哪里错了...不好意思 打错了

来源:学生作业帮助网 编辑:作业帮 时间:2024/09/02 23:17:33
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